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Chapter 2 – The Basic FEA Procedure
Now, equations 2.2, 2.4, 2.5 also contain the u
2
term and therefore these equations must
also be modified. We can modify equation 1 by transferring the term k
12
u
2
to the right
hand side and replacing u
2
by U
2
. The modified equation can be written as
K
11
u
1
+ 0 + k
13
u
3
+ k
14
u
4
= F
1
– k
12
U
2
Similarly, equations 3 and 4 can be written as
K
31
u
1
+ 0 + k
33
u
3
+ k
34
u
4
= F
3
– k
32
U
2
K
41
u
1
+ 0 + k
43
u
3
+ k
44
u
4
= F
4
– k
42
U
2
The final matrix equation is
k
11
0
k
13
k
14
u
1
F
1
– k
1
U
2
0 k
22
0 0
u
2
= k
22
U
2
k
31
0
k
33
k
34
u
3
F
3
– k
32
U
2
k
41
0
k
43
k
44
u
4
F
4
– k
42
U
2
The dotted line indicates changes made in the enclosed terms. The final matrix remains
symmetric and has the same size. The boundary conditions for forces can now be
incorporated and a numerical solution scheme can be used to solve this equation. This
procedure is summarized in the following simple, step-by-step approach.
Given the known boundary conditions at node 2: u
i
= u
2
= U
2
, follow these steps to
incorporate the known nodal values. Note that, here, i = 2 and j = 1,2,3,4.
Step 1: Set all terms in row 2 to zero, except the term in column 2 (kij = 0, k
ii
= k
22
≠
0)
Step 2: Replace F
2
with the term k
22
U
2
(F
i
= k
ii
u
i
)
Step 3: Subtract the value k
i2
U2 from all the forces, except F
2
( subtract k
ji
from the
existing values of f
j
), where i = 1, 3, and 4
Step 4: Set all the elements in column 2 to zero, except, row2 (all k
ji
= 0, k
ii
# 0)
The above procedure now will be applied in the following example problem.
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-16
Chapter 2 – The Basic FEA Procedure
Example 2.5
In example problem 2.4 replace the force F by a nodal deflection of 1.5 mm on node 2
and rework the problem.
Solution
Rewriting the final structural matrix equation in example 2.4, we have
F
1
25 -15 0 -10 u
1
F
2
= -15 90 -45 -30 u
2
F
3
0 -45 80 -35 u
3
F
4
-10 -30 -35 75 u
4
Boundary condition are: u
1
= u
4
= 0, and u
2
= U
2
= 1.5mm. Applying the 4 steps
described above in sequence,
Step 1: Set all terms in row 2 to zero, except the term in column 2 (kij = 0, k
ii
= k
22
≠
0)
F
1
25 -15 0 -10
u
1
F
2
= 0 90 0 0
u
2
F
3
0 -45 80 -35
u
3
F
4
-10 -30 -35 75
u
4
Step 2: Replace F
2
with the term k
22
U
2
= (90)(1.5) = 135, (F
i
= k
ii
u
i
)
F
1
25 -15 0 -10
u
1
135 = 0 90 0 0
u
2
F
3
0 -45 80 -35
u
3
F
4
-10 -30 -35 75
u
4
Step 3: Subtract the value k
22
U
2
from all the forces, except F
2
(subtract k
ji
from the
existing values of f
j
)
F
1
Æ
F
1
– (15)(1.5) = 22.5 Row 1: k
j2 =
k
12 =
-15
F
3
Æ
F
3
– (-45)(1.5) = 67.5
Row 2: k
j2 =
k
32 =
-45
F
4
Æ
F
4
– (-30)(1.5) = 45
Row 2: k
j2 =
k
42 =
- 30
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-17
Chapter 2 – The Basic FEA Procedure
Note: F
1
= F
3
= F
4
= 0.
The new force equation now is,
22.5
135
67.5
45
Step 4: Set all the elements in column 2 to zero, except, row2 (all k
ji
= 0, k
ii
≠
0)
Or, k
12
= k
32
= k
42
= 0, and the new equation is,
22.5
25 0 0 -10 u
1
135 = 0 90 0 0 u
2
67.5
0 0 80 -35 u
3
45 -10 0 -35 75 u
4
This is the final equation after the nodal value u
2
= 1.5 mm is incorporated into the
structural equation.
The same procedure can be followed for the boundary conditions u
1
= u
4
= 0. It can be
stated that for zero nodal values, the procedure will always lead to elimination of rows
and columns corresponding to these nodes, that is, the first and fourth rows as well as
columns will drop out. The reader is encouraged to verify this statement.
Thus, the final equation is,
90 0 u
2
= 135
0 80 u
3
67.5
Solving for u
2
and u
3
, we get
u
2
= 1.5
u
3
0.8437
Spring deflection is:
Spring 1:
u
2
– u
1
= 1.500
Spring 2: u
3
– u
1
= 0.8437
Spring 3: u
3
– u
2
= -0.6563
Spring 4: u
3
– u
2
= -0.6563
Spring 5:
u
4
– u
2
= -1.500
Spring 6:
u
4
– u
3
= -1.6875
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-18
Chapter 2 – The Basic FEA Procedure
2.3.6 Structures That can be Modeled Using a Spring Elements
As mentioned earlier, almost all engineering materials are similar to a linear spring,
satisfying the relation F = ku. Therefore, any structure that deflects only along its axial
direction (with one degree of freedom) can be modeled as a spring element. The
following example illustrates this concept.
Example 2.6
A circular concrete beam structure is loaded as shown. Find the deflection of points at 8”,
16”, and the end of the beam. E = 4 x 106 psi
y
12 in
3 in 50000 lb
x
24 in
Figure 2.7
Solution
The beam structure looks very different from a spring. However, its behavior is very
similar. Deflection occurs along the x-axis only. The only significant difference between
the beam and a spring is that the beam has a variable cross-sectional area. An exact
solution can be found if the beam is divided into an infinite number of elements, then,
each element can be considered as a constant cross-section spring element, obeying the
relation F = ku, where k is the stiffness constant of a beam element and is given by k =
AE/L.
In order to keep size of the matrices small (for hand- calculations), let us divide the beam
into only three elements. For engineering accuracy, the answer obtained will be in an
acceptable range. If needed, accuracy can be improved by increasing the number of
elements.
As mentioned earlier in this chapter, spring, truss, and beam elements are line-elements
and the shape of the cross section of an element is irrelevant. Only the cross-sectional
area is needed (also, moment of inertia for a beam element undergoing a bending load
need to be defined). The beam elements and their computer models are shown in figure
2.8.
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-19
Chapter 2 – The Basic FEA Procedure
Here, the question of which cross-sectional area to be used for each beam section arises.
A good approximation would be to take the diameter of the mid-section and use that to
approximate the area of the element.
k
1
k
2
k
3
k
1
k
2
k
3
1 2 3
1
2 2
3 3 4
1
2
3 4
Beam sections
Equivalent spring elements
Figure 2.8
Cross-sectional area
The average diameters are: d
1
= 10.5 in., d
2
= 7.5 in., d
3
= 4.5. (diameters are taken at the
mid sections and the values are found from the height and length ratio of the triangles
shown in figure 2.10), which is given as
12/L = 3/(L-24),
L = 32
Average areas are:
A
1
= 86.59 in2
A
2
= 56.25 in2
A
3
= 15.9 in2
24 in
12 in d
1
d
2
d
3
3 in
Original Averaged
8 8 8
L- 24
L
Figure
2.9
Figure
2.10
Stiffness
k
1
= A
1
E/L
1
= (86.59)(4
×
10
6
/8) = 4.3295
×
10
7
lb./in., similarly,
ME 273 Lecture Notes © by R. B. Agarwal
Finite Element Analysis
2
-20